Problem: Let $R$ be the region enclosed by the polar curve $r(\theta)=2\theta\sin(\theta)$, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{2\pi}4\theta^2\sin^2(\theta)\,d\theta$ (Choice B) B $ \int_{0}^{2\pi}2\theta^2\sin^2(\theta)\,d\theta$ (Choice C) C $ \int_{0}^{\pi}2\theta^2\sin^2(\theta)\,d\theta$ (Choice D) D $ \int_{0}^{\pi}4\theta^2\sin^2(\theta)\,d\theta$
Explanation: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. $\alpha$ is the first non-negative $\theta$ -value for which $r(\theta)=0$. That means $\alpha=0$. $\beta$ is the next $\theta$ -value after $\alpha$ for which $r(\theta)=0$. That means $\beta=\pi$. Let's plug ${r(\theta)=2\theta\sin(\theta)}$, ${\alpha=0}$, and ${\beta=\pi}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{0}}^{{\pi}}\dfrac{1}{2}\left({2\theta\sin(\theta)}\right)^{2}d\theta \\\\ &= \int_{0}^{\pi}\dfrac{1}{2}\cdot4\theta^2\sin^2(\theta)\,d\theta \\\\ &= \int_{0}^{\pi}2\theta^2\sin^2(\theta)\,d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{0}^{\pi}2\theta^2\sin^2(\theta)\,d\theta$